POJ3624 0-1 Backpack (dp+ scrolling array)

Charm Bracelet

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 47440		Accepted: 20178

Description

Bessie have gone to the mall ' s jewelry store and spies a charm bracelet. Of course, she ' d like-to-fill it with the best charms possible from the N (1≤ n ≤3,402) available Char Ms. Each charm I in the supplied list has a weight wi (1≤ wi ≤400), a ' desirability ' factor C5>di (1≤ Di ≤100), and can be used at the most once. Bessie can only support a charm bracelet whose weight are no more than m (1≤ m ≤12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the MA Ximum possible sum of ratings.

Input

* Line 1:two space-separated integers: N and M
* Lines 2. N+1:line i+1 describes charm I with II space-separated integers: Wi and Di

Output

* Line 1: A single integer which is the greatest sum of charm desirabilities so can be achieved given the weight Constrai Nts

Sample Input

4 61 42 63 122 7

Sample Output

23

If using Dp[n][m]: The first n items do not exceed the maximum value of volume m, the state transfer equation is:

DP[I][J] = 0 (i==0orj==0)//no items, backpack volume of the largest value is also 0, and more items, backpack volume of 0 are not fit.

DP[I][J]=DP[I-1][J] (J<d[i])//Current backpack plug no more than the first item, the maximum value and the first i-1 items not greater than the maximum value of volume J is the same.

Dp[i][j]=max (Dp[i-1][j],dp[i-1][j-d[i]]+w[i])//Current backpack If you can load the first item, then the maximum value after loading the article I and the maximum value of the item I do not fit the maximum value

However, because the range of N and M is too large, open two-dimensional array will burst memory, the analysis solution found that the solution Dp[i][j] is only a row above it and a value to the left of the previous row, that is, only the row above it, you can use a scrolling one-dimensional array solution, the new value to the upper position, Note, however, that the order must be right-to-left, or it will overwrite useful values.

1#include <iostream>2#include <cstring>3#include <algorithm>4 using namespacestd;5 intMain ()6 {7     intN, M;8     intd[13010], w[13010];9     intdp[13010];//since solving the row is only related to the previous row, you can use a scrolling array, dp[j] to indicate the maximum value of the first I item without exceeding the volume JTenCIN >> N >>m; One      for(inti =1; I <= N; ++i) A     { -CIN >> D[i] >>W[i]; -     } theMemset (DP,0,sizeof(DP)); -      for(inti =1; I <= N; ++i) -     { -          for(intj = m; J >=1; --J)//solve from right to left, save the newly calculated value in the upper position, because the overridden value is only relevant to the right of the next and lower line +         { -             if(J >=D[i]) +DP[J] = max (Dp[j], Dp[j-d[i]] +w[i]); A             Else       //if j-d[i]<0, which indicates that the capacity of the current backpack is not enough to put the first item, that the maximum value of the pre-I item with the maximum value of the pre-i-1 goods is equal, at                  Break;//also because the left side of J is only smaller because it is evaluated from right to left, so you can skip saving time -                          -         } -     } -cout << Dp[m] <<Endl; -     return 0; in}

POJ3624 0-1 Backpack (dp+ scrolling array)